Exponential inequality solution

Question

I am trying to solve $$ \displaystyle (2^{(3x-1)/(x-1)})^{1/3} < 8^{((x-3)/(3x-7))}$$

Here's how I proceeded.

$LHS = 2^{((3x-1)/(3(x-1))} , RHS = 2^{((3*(x-3))/(3x-7))}$

hence inequating the exponents ,

$$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$

$$ \implies (3x-1)(3x-7) < 9(x-1)(x-3)$$

$$\implies 9x^2-24x+7 < 9x^2-36x+27$$

cancelling terms

$$12x <20$$

$$=> x < 5/3 $$

but the book states the solution as $(-\infty , 1)$ U $(5/3,7/3)$

Not sure how this has been arrived at , can someone help me understand ?

Thanks in advance.

Madavan.

Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. — N. F. Taussig, Jun 09 '18 at 13:13
When you multiply an inequality you have to care about the sign of the factor... there are many cases...negatif or positif... — user577215664, Jun 09 '18 at 13:27

user577215664 · Answer 1 · 2018-06-09T15:09:21.280

1

Be carefull when you multiply your inequality $$\frac {(3x-1)}{(3(x-1))} < 3\frac {(x-3)}{(3x-7)}$$ Because $3x-7$ or $3(x-1)$ may be negative and change the order of the ineqaulity

You need to study the sign of $3(x-1)(3x-7)$...

A)$ \,3(x-1)(3x-7) $ is positive

$$\implies (3x-1)(3x-7) < 9(x-1)(x-3)$$ $$ \implies x<5/3$$

First case $3(x-1)>0$ and $3x-7>0 \implies x> \frac 73$ With the constraint $x>7/3$ there are no solution.

Second case $3(x-1)<0$ and $3x-7<0$

$$\implies x<1$$

with this constraint ($x<1$) you get a part of the result of the book $ \implies x \in [-\infty,1]$.

Now you have to study the case when $3(x-1)(3x-7)$ is negative to get the other part of the solution of the book...

B)$ \,3(x-1)(3x-7) $ is negative

$$ (3x-1)(3x-7) > 9(x-1)(x-3)$$ $$ \implies x>\frac 5 3$$

First case $3(x-1)<0 \quad ; 3x-7>0 \implies x<1, x>7/3$ Impossible ..

Second case $3(x-1)>0 \quad ; 3x-7<0 \implies x>1, x<7/3$ .

With the constraint you finally get $x \in (5/3,7/3)$

And the complete solution is $(-\infty,1)$ U $(5/3,7/3)$

edited Jun 09 '18 at 15:09

answered Jun 09 '18 at 13:28

user577215664

40,625

Sorry i dont understand the reply fully , from where does the 5/3 come in to picture ? Can you please explain in more detail ? Thanks. – Madavan Viswanathan Jun 09 '18 at 14:28
First consider that both polynomials are positive you get a constraint on the variable x ...When you solved the inequality you get some values for x...you check that both the solution and the constraint match @Madavan – user577215664 Jun 09 '18 at 14:31
@Madavan whern you multiply an inequality by a positive number it stay the same. For example multily by 3 $$x>3 \implies 3x>9$$ But when the number is negative it changes the inequality. Example multiply by -3 $$x>3 \implies -3x<-9$$ – user577215664 Jun 09 '18 at 14:38
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Got it , thanks very much Isham. – Madavan Viswanathan Jun 09 '18 at 15:07
yw @Madavan ... – user577215664 Jun 09 '18 at 15:12

Exponential inequality solution

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